AitkattlercarlyF
contestada

In major league baseball, the pitcher's mound is 60 feet from the batter.
If a pitcher throws a 91 mph fastball, how much time elapses from when the ball leaves the pitcher's hand until the ball reaches the batter?

Respuesta :

scme1702
Converting the speed from miles per hour to feet per second:
91 mph = 133.5 feet / second

Time = distance / speed
Time = 60 / 133.5
Time = 0.45 seconds

Answer: 0.48 s

Distance covered, [tex]d= 60 ft [/tex]

In SI units,

[tex]1 ft = 0.3048 m[/tex]

[tex] 60 ft= 60 ft \times 0.3048 m/ft=18.28 m[/tex]

Speed, [tex] v= 91 mph= 91 miles/hr \times 1.61 km/miles \times \frac{5}{18}\frac {m/s}{km/hr}=38.16 m/s}[/tex]

[tex]Time elapsed= \frac{Distance}{speed}[/tex]

[tex]t=\frac{18.28 m}{38.16 m/s}=0.48 s[/tex]




Otras preguntas